MATHEMATICS

A. I. Kokorin

On Order-Extendable Groups

(Presented by Academician A. I. Mal'tsev on 24 I 1963)

In this note one finds subgroups of an order-extendable group, definable independently of the order, by which the factor-groups are order-extendable. In particular, these subgroups are of interest in connection with questions of describing the ways of ordering a group. Next, the connection is indicated between the problem of the coincidence of the classes of order-extendable and orderable groups and other questions.

A group is called order-extendable if every partial order on it can be extended to a linear one.

Notation: \(S(a,b,\ldots)\) is the minimal invariant semigroup in the group \(G\) containing \(a,b,\ldots\); \(\overline{M}\) is the image of the set \(M\) from the group \(G\) under a homomorphism of \(G\) onto the group \(\overline{G}\).

A subgroup \(H\) of a group \(G\) is called strictly isolated if from
\[ xg_1^{-1}xg_1\ldots g_n^{-1}xg_n \in H \]
it follows that
\[ x,\ g_1^{-1}xg_1,\ldots,\ g_n^{-1}xg_n \in H. \]
Subgroups and subsemi-groups of this kind were considered in \((^1,^2)\).

Theorem 1. The factor-group \(G/H\) of an order-extendable group \(G\) is order-extendable if and only if \(H\) is an invariant strictly isolated subgroup.

The necessity follows from the strict isolation of the identity in an ordered group. To prove sufficiency, suppose at first that \(H\) is strictly isolated and \(G/H=\overline{G}\) is not orderable. Then, on the basis of Theorem 1 from \((^1)\), in \(\overline{G}\) there are elements \(\overline{a}_1,\ldots,\overline{a}_n\), distinct from \(\overline{1}\), such that for arbitrary \(\varepsilon_i=\pm1\)

\[ S(\overline{a}_1^{\varepsilon_1},\ldots,\overline{a}_n^{\varepsilon_n})\cap \overline{1}\ne\phi^* \]

and, consequently, in \(G\) also

\[ S(a_1^{\varepsilon_1},\ldots,a_n^{\varepsilon_n})\cap H\ne\phi. \tag{*} \]

Here one may assume \(n\) minimal, i.e., no proper part of \(a_1,\ldots,a_n\) has this property. Let, further, \(P\) be the semigroup of positive elements (without the identity) from \(H\) under some ordering of \(G\), and
\[ P_1=S(a_1)H\cup P. \]
\(P_1\) is an invariant semigroup by virtue of the invariance of \(H\), \(S(a_1)\), and \(P\). We shall show that \(P_1\) does not contain the identity. If \(s\in S(a_1)\), \(h\in H\), and \(sh=1\), then \(s\in H\), and, by the strict isolation of \(H\), it follows that \(a_1\in H\), which contradicts \(\overline{a}_1\ne\overline{1}\). Thus \(P_1\) is an invariant semigroup not containing the identity and, hence, it defines in \(G\) a partial order. Starting from \(P_1\), extend the order of \(G\). We shall assume \(a_1,\ldots,a_n>1\), since if \(a_i<1\), one may take \(a_i^{-1}\). Moreover, for every \(h\in H\) we shall have \(a_1\gg h\) (i.e. \(a_1>h^k\) for every integer \(k\)), since \(a_1h^{-k}\in S(a_1)H\subset P_1\). Next take the intersection \(H_1\) of all convex subgroups containing \(H\). Let
\[ a_1,\ldots,a_m\in H_1,\quad a_{m+1},\ldots,a_n\notin H_1. \]
\(m\ge 1\), since from \(a_1\in H\) it follows that \(a_1\in H_1\). On the basis of convexity and invariance of \(H_1\), for any \(g\in G\), \(h\in H_1\) we shall have
\[ g^{-1}a_1g,\ldots,g^{-1}a_mg\gg h, \]
and hence
\[ g^{-1}a_1g,\ldots,g^{-1}a_mg\in H_1. \]
Therefore every product composed of elements of \(H_1\) and at least one of
\[ g^{-1}a_1g,\ldots,g^{-1}a_mg \]
also does not belong to \(H_1\), and hence not to \(H\).

\(*\) \(\phi\) is the empty set.

Consequently,

\[ S(a_1,\ldots,a_n)\cap H=S(a_{m+1},\ldots,a_n)\cap H\ne\varnothing. \]

But, by the minimality of \(n\), there exist such \(\varepsilon_{m+1},\ldots,\varepsilon_n\) that

\[ S(a_{m+1}^{\varepsilon_{m+1}},\ldots,a_n^{\varepsilon_n})\cap H\ne\varnothing, \]

and, therefore,

\[ S(a_1,\ldots,a_m,a_{m+1}^{\varepsilon_{m+1}},\ldots,a_n^{\varepsilon_n})\cap H=\varnothing, \]

which contradicts \((*)\). Hence \(G/H\) is orderable.

We now show that \(G/H=\overline G\) is an orderable group. Let \(\overline P_2\) be the semigroup of positive elements of \(\overline G\) under some partial ordering of \(\overline G\), and let \(P_2\) be its complete preimage in \(G\). Starting from the partial order defined by \(P_2\), we order \(G\). In this case, from \(P_2=P_2H\) it follows that \(P_2\cap H_2=\varnothing\), where \(H_2\) is the intersection of all convex subgroups containing \(H\). On the basis of \((^3)\), from the orderability of \(G/H_2\) and the possibility of ordering \(H_2/H\), which is preserved under the action of inner automorphisms from \(G/H\), the group \(\overline G\) can be ordered while preserving the orders on \(G/H_2\) and on \(H_2/H\). This order will be an extension of \(\overline P_2\), since \(P_2\cap H=\varnothing\).

Corollary. If in an orderable group \(G\) a subgroup \(H\) is invariant and strictly isolated, then there exists an ordering of \(G\) under which \(H\) is convex.

Lemma. If in an ordered group \(G\): 1) an element \(z\) belongs to the center of the group \(G\); 2) \(A\) is the union of all convex subgroups not containing \(z\); 3) \(xg_1^{-1}xg_1\ldots g_n^{-1}xg_n=z\), then \(\overline{x}^{\,n+1}=\overline z\) in \(\overline G=G/A\).

Let \(B\) be the intersection of all convex subgroups containing \(z\). From 1), 2), and the construction of \(B\) it follows that \(A\) and \(B\) are invariant. By convexity and invariance of \(A\) and \(B\) we have \(x,g_1^{-1}xg_1,\ldots,g_n^{-1}xg_n\in B\setminus A\). The inner automorphisms of the group \(G\) induce in \(B/A\) automorphisms that are order-preserving (see \((^4)\)). Therefore, for any \(h\in G\) we have

\[ \overline z=\overline{h^{-1}zh} =\overline{h^{-1}xh\cdot h^{-1}g_1^{-1}xg_1h\ldots h^{-1}g_n^{-1}xg_nh} \]

\[ =\overline{(h^{-1}xh)\cdot g_1^{-1}(h^{-1}xh)g_1\ldots g_n^{-1}(h^{-1}xh)g_n}. \]

Suppose, for definiteness, that \(\overline x>\overline{h^{-1}xh}\) in \(\overline G\). But then

\[ \overline{xg_1^{-1}xg_1\ldots g_n^{-1}xg_n}> \]

\[ >\overline{(h^{-1}xh)g_1^{-1}(h^{-1}xh)g_1\ldots g_n^{-1}(h^{-1}xh)}. \]

Consequently, \(\overline x=\overline{h^{-1}xh}\), and therefore \(\overline{x}^{\,n+1}=\overline z\) in \(\overline G\).

Theorem 2. The quotient group \(G/Z\) of an orderable group \(G\) by its center \(Z\) is orderable.

For the proof, by Theorem 1, it is enough to show that the center \(Z\) is strictly isolated. Order \(G\), and let \(z\in Z\); \(z\in B\setminus A\), where \(A\) and \(B\) are neighboring convex subgroups; \(P\) is the set of positive elements of \(G\) not belonging to \(B\setminus A\); \(I^+(z)\) is the set of positive elements of \(I(z)\), where \(I(z)\) is the isolator (see \((^5)\)) of \(z\) in \(G\). By the isomorphism of \(I(z)\) to an additive group of rational numbers and by the Archimedean property of \(B/A\), it follows that \(I(z)\setminus 1\subseteq B\setminus A\). From this, convexity and invariance of \(A\) and \(B\) imply that

\[ P_1=P I(z)\cup I^+(z) \]

is an invariant semigroup not containing the identity, and hence \(P_1\) defines a partial order in \(G\). Starting from \(P_1\), order \(G\). Let under this ordering \(H\) be the intersection of all convex subgroups containing \(z\). Then \(H\setminus 1\subseteq B\setminus A\), since, if \(g\in B\setminus A\),

then either \(g \in P\), or \(g \in P^{-1}\). We may further assume that the order in the subgroup \(H\) coincides with its order under the original ordering of \(G\). From this, \(H\setminus 1 \subseteq B\setminus A\), and the Archimedeanness of \(B/A\), it follows that \(H\) is Archimedean and is the least convex subgroup distinct from the identity under the new ordering of \(G\). Therefore, on the basis of the lemma, the equality
\[ x g_1^{-1} x g_1 \cdots g_n^{-1} x g_n = z \]
implies \(x^{n+1}=z\) in \(G/\{1\}=G\). From \(x^{n+1}=z\) and the isolatedness of \(I(z)\) it follows that \(x \in I(z)\subseteq Z\), and hence \(g_1^{-1}xg_1,\ldots,g_n^{-1}xg_n\in Z\). The theorem is proved.

Remark 1. From the isomorphism of an ordered group to a factor group of some free ordered group (see \((^4)\)), Theorems 1 and 2 imply the equivalence of the following propositions.

1) An orderable group is bi-orderable.

2) A free group is bi-orderable.

3) A group is bi-orderable if and only if its identity is strictly isolated.

4) The factor group of a group with strictly isolated identity by the center is bi-orderable.

Remark 2. A two-step solvable group is orderable if and only if its identity is strictly isolated.

This is proved by a direct application of Theorem 3 from \((^1)\) to the system consisting of the group itself, the isolator of its commutant, and the identity.

Remark 3. The factor group \(G/Z_1\) of an orderable group \(G\) by a complete subgroup \(Z_1\) of the center \(Z\) is orderable.

For the proof, take the system of all convex subgroups under some ordering of \(G\):
\[ G \supset \cdots \supset B \supset A \supset \cdots \supset C \supset \cdots \supset 1 \tag{**} \]
and multiply each subgroup of this system by \(Z_1\). Then, together with the identity, we obtain the system
\[ G \supset \cdots \supset Z_1B \supset Z_1A \supset \cdots \supset Z_1C \supset \cdots \supset Z_1 \supset 1, \tag{***} \]
which, as we shall show, satisfies the conditions of Theorem 3 from \((^1)\). This will prove the assertion, since under the order constructed in \((^1)\), the subgroups of the system \((***)\) will be convex. We shall show the validity of only one of these conditions (the strict isolatedness of the subgroups of the system \((***)\)), since the fulfillment of the remaining ones follows directly from their fulfillment for the system \((**)\).

Let
\[ x g_1^{-1} x g_1 \cdots g_n^{-1} x g_n = zc \in Z_1C, \]
where \(z\in Z_1\), \(c\in C\), \(z\in C\) and \(z\in B\setminus A\), where \(A\) and \(B\) are adjacent convex subgroups of \(G\). Then, by the lemma, in \(G/A\) we have \(\overline z=\overline x^{\,n+1}\), and hence \(z=x^{n+1}a\), where \(a\in A\). By the completeness of \(Z_1\),
\[ (z_1x^{-1})^{n+1}=a, \]
where \(z_1^{\,n+1}=z\) and \(z_1\in Z_1\). From the strict isolatedness of \(A\), \((z_1x^{-1})^{n+1}=a\) implies \(z_1x=a_1^{-1}\in A\), i.e. \(x=z_1a_1\).

Further, from
\[ zc=(z_1a_1)g_1^{-1}(z_1a_1)g_1\cdots g_n^{-1}(z_1a_1)g_n \]
it follows that
\[ a_1 g_1^{-1}a_1 g_1\cdots g_n^{-1}a_1 g_n=c\in C, \]
whence, on the basis of the strict isolatedness of \(C\),
\[ a_1,\ g_1^{-1}a_1g_1,\ldots,\ g_n^{-1}a_1g_n\in C, \]
and therefore \(x=z_1a_1\in Z_1C\). Consequently,
\[ x,\ g_1^{-1}xg_1,\ldots,\ g_n^{-1}xg_n\in Z_1C. \]

In conclusion I express my gratitude to the supervisor of the work, P. G. Kontorovich.

Received
15 I 1963

CITED LITERATURE

1. V. D. Podderyugin, Izv. AN SSSR, Ser. Mat., 21, No. 2, 199 (1957).
2. A. I. Mal’cev, Collected volume Algebra and Logic, Seminar, 1, issue 2, Novosibirsk, 1962, p. 5.
3. E. P. Shimbireva, Mat. sbornik, 20 (62), 145 (1947).
4. K. Iwasawa, J. Math. Soc. Japan, 1, 1 (1948).
5. P. G. Kontorovich, Mat. sbornik, 22, 79 (1948).